Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        Consider a wire of length 4 m and cross-sectional area 1 mm^2 carrying a current 2 A. If each cubic metre of the material contains 10^29 free electrons, find the average time taken by an electron to cross the length of the wire.`
3 years ago

Deepak Patra
474 Points
```										Sol. ℓ = 4 m, A = 1 mm^2 = 1 * 10^–6 m^2
I = 2 A, n/V = 10^29, t = ?
i = n A V base d e
⇒ e = 10^29 * 1 * 10^–6 * V base d * 1.6 * 10^–19
⇒ V base d = 2/10^29 *10^-6 *1.6 *10^-19
= 1/0.8 * 10^4 = 1/8000
t = ℓ/V base b = 4/1/8000 = 4 * 8000
= 32000 = 3.2 * 10^4 sec.

```
3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Electromagnetic Induction

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Electricity and Magnetism
• OFFERED PRICE: Rs. 1,696
• View Details